Find the Maximum/Minimum Value f(x)=-4x^2+48x-149

$f (x) = - 4 x^{2} + 48 x - 149$

Step 1

The maximum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is negative, the maximum value of the function is $f (- \frac{b}{2 a})$ .

$f_{max}$ $x = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a}$

Step 2

Find the value of $x = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$x = - \frac{48}{2 (- 4)}$

Remove parentheses.

$x = - \frac{48}{2 (- 4)}$

Simplify $- \frac{48}{2 (- 4)}$ .

Cancel the common factor of $48$ and $2$ .

Factor $2$ out of $48$ .

$x = - \frac{2 \cdot 24}{2 \cdot - 4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 4$ .

$x = - \frac{2 \cdot 24}{2 (- 4)}$

Cancel the common factor.

$x = - \frac{2 \cdot 24}{2 \cdot - 4}$

Rewrite the expression.

$x = - \frac{24}{- 4}$

Cancel the common factor of $24$ and $- 4$ .

Factor $4$ out of $24$ .

$x = - \frac{4 (6)}{- 4}$

Move the negative one from the denominator of $\frac{6}{- 1}$ .

$x = - (- 1 \cdot 6)$

Multiply $- (- 1 \cdot 6)$ .

Multiply $- 1$ by $6$ .

$x = - - 6$

Multiply $- 1$ by $- 6$ .

$x = 6$

Step 3

Evaluate $f (6)$ .

Replace the variable $x$ with $6$ in the expression.

$f (6) = - 4 {(6)}^{2} + 48 (6) - 149$

Simplify the result.

Simplify each term.

Raise $6$ to the power of $2$ .

$f (6) = - 4 \cdot 36 + 48 (6) - 149$

Multiply $- 4$ by $36$ .

$f (6) = - 144 + 48 (6) - 149$

Multiply $48$ by $6$ .

$f (6) = - 144 + 288 - 149$

Simplify by adding and subtracting.

Add $- 144$ and $288$ .

$f (6) = 144 - 149$

Subtract $149$ from $144$ .

$f (6) = - 5$

The final answer is $- 5$ .

$- 5$

Step 4

Use the $x$ and $y$ values to find where the maximum occurs.

$(6, - 5)$

Step 5

Find the Maximum/Minimum Value f(x)=-4x^2+48x-149

Related Questions

Support LCMGCF.com by disabling your adblocker.